In: Statistics and Probability
How to Run descriptives and a two-tailed, two sample assuming equal variance t-test.
Here's your data:
| Weight of Apples in Grams | |||
| Apple ID | Farm A | Farm B | |
| 1 | 131 | 151 | |
| 2 | 147 | 159 | |
| 3 | 134 | 162 | |
| 4 | 134 | 158 | |
| 5 | 136 | 159 | |
| 6 | 137 | 160 | |
| 7 | 140 | 150 | |
| 8 | 134 | 160 | |
| 9 | 136 | 160 | |
| 10 | 133 | 160 | |
| 11 | 134 | 160 | |
| 12 | 132 | 158 | |
| 13 | 139 | 162 | |
| 14 | 136 | 160 | |
| 15 | 135 | 154 | |
| 16 | 135 | 155 | |
| 17 | 135 | 159 | |
| 18 | 138 | 151 | |
| 19 | 134 | 151 | |
| 20 | 149 | 150 | |
| 21 | 149 | 155 | |
| 22 | 135 | 150 | |
| 23 | 148 | 150 | |
| 24 | 135 | 158 | |
| 25 | 149 | 152 | |
| 26 | 137 | 162 | |
| 27 | 134 | 158 | |
| 28 | 140 | 161 | |
| 29 | 138 | 160 | |
| 30 | 133 | 161 | |
| 31 | 140 | 158 | |
| 32 | 138 | 160 | |
| 33 | 150 | 149 | |
| 34 | 135 | 160 | |
| 35 | 136 | 160 | |
| 36 | 148 | 157 | |
| 37 | 150 | 152 | |
| 38 | 136 | 150 | |
| 39 | 142 | 157 | |
| 40 | 132 | 149 | |
| 41 | 133 | 160 | |
| 42 | 147 | 150 | |
| 43 | 135 | 159 | |
| 44 | 140 | 153 | |
| 45 | 132 | 151 | |
| 46 | 132 | 160 | |
| 47 | 136 | 151 | |
| 48 | 141 | 162 | |
| 49 | 136 | 162 | |
| 50 | 135 | 160 | |
| 51 | 142 | 160 | |
| 52 | 135 | 150 | |
| 53 | 132 | 158 | |
| 54 | 140 | 160 | |
| 55 | 136 | 152 | |
| 56 | 138 | 160 | |
| 57 | 132 | 159 | |
| 58 | 136 | 150 | |
| 59 | 136 | 161 | |
| 60 | 135 | 158 | |
| 61 | 135 | 152 | |
| 62 | 136 | 154 | |
| 63 | 137 | 151 | |
| 64 | 138 | 150 | |
| 65 | 137 | 150 | |
| 66 | 136 | 153 | |
| 67 | 134 | 153 | |
as the data is large enough i am using MINITAB to solve the problem.
steps:-
copy the data in minitab 
 stat
 basic
statistics 
 2 sample
t  
 choose each
sample in its own column 
 in sample 1
select 'farm A' and in sample 2 select 'farm B' 
 click
options 
 type 95 in
confidence interval 
 type 0.0 in
hypothesized difference 
 choose the
alternative hypothesis difference 
 hypothesized
difference 
 tick assume
equal variance 
 ok
 ok.
your output be :-
Two-Sample T-Test and CI: Farm A, Farm B
Method
| μ₁: mean of Farm A | 
| µ₂: mean of Farm B | 
| Difference: μ₁ - µ₂ | 
Equal variances are assumed for this analysis.
Descriptive Statistics
| Sample | N | Mean | StDev | SE Mean | 
| Farm A | 67 | 137.55 | 5.02 | 0.61 | 
| Farm B | 67 | 156.07 | 4.43 | 0.54 | 
Estimation for Difference
| Difference | Pooled StDev  | 
95% CI for Difference  | 
| -18.522 | 4.735 | (-20.141, -16.904) | 
Test
| Null hypothesis | H₀: μ₁ - µ₂ = 0 | 
| Alternative hypothesis | H₁: μ₁ - µ₂ ≠ 0 | 
| T-Value | DF | P-Value | 
| -22.64 | 132 | 0.000 | 
decision:-
p value = 0.000 <0.05
so, we reject the null hypothesis.
we conclude that,
there is sufficient evidence to claim that there is a difference between weight of apples of farm A and farm B at 0.05 level of significance.
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